Strength of Suspension Link.
405
Example 10.—Find the thickness of a short, hollow, cast-iron
column of 18" outside diameter, to sustain a live load of 80 tons, plus
a dead load of 100 tons. (Eng\ Exam. 1888.)
Equivalent dead load = 100 + (2 x 80) = 260 tons
260 = 4 a and a = 65 sq. ins.
But<2=7rR2-7r r2=65 or — (8i-^2)=6s
andr= V 60*3 =7*8 .'. /=9 -7*8 = 1 "z ins.
Shear Stress-Action rarely occurs alone, but pins and rivets
are thus calculated : W = /s #. *
Strength of a Suspension Link (see Fig. 355).—The
strength of one thin link in tension, at a and c\ the shear
strength of the pin d-} the strength at b] and the bearing stress
on projected area of e, should each equal half the load :
(i) (2) (3)* (4) (5)
(w -
Let /t-i, /b=4, and/,-}.
By(4>and(5)
By (2) and (4)
By (3) and (4) |x^~i*/
By actual tests ^ =/
and* the thick link must be 2 / in thickness.
and
'66^
i Kbt and w=:
and / = '20 &
«. '66^
p. 415.
(See Appendix IV., #. 952.)